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Sum of a(n) terms of 1/k^(8/9) first exceeds n.
0

%I #2 Mar 30 2012 17:30:25

%S 1,2,4,9,18,37,70,128,225,383,633,1019,1601,2461,3711,5494,8003,11484,

%T 16249,22698,31328,42762,57764,77274,102436,134634,175532,227123,

%U 291776,372300,472004

%N Sum of a(n) terms of 1/k^(8/9) first exceeds n.

%t s = 0; k = 1; Do[ While[ s <= n, s = s + N[ 1/k^(8/9), 24 ]; k++ ]; Print[ k - 1 ], {n, 1, 30} ]

%Y Cf. A019529 and A002387.

%K nonn

%O 0,2

%A _Robert G. Wilson v_, Aug 01 2000